Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z + 1}{z^2 + 14z + 45} \times \dfrac{z^2 + 9z}{z + 1} $
Solution: First factor the quadratic. $r = \dfrac{z + 1}{(z + 9)(z + 5)} \times \dfrac{z^2 + 9z}{z + 1} $ Then factor out any other terms. $r = \dfrac{z + 1}{(z + 9)(z + 5)} \times \dfrac{z(z + 9)}{z + 1} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (z + 1) \times z(z + 9) } { (z + 9)(z + 5) \times (z + 1) } $ $r = \dfrac{ z(z + 1)(z + 9)}{ (z + 9)(z + 5)(z + 1)} $ Notice that $(z + 1)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ z(z + 1)\cancel{(z + 9)}}{ \cancel{(z + 9)}(z + 5)(z + 1)} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $r = \dfrac{ z\cancel{(z + 1)}\cancel{(z + 9)}}{ \cancel{(z + 9)}(z + 5)\cancel{(z + 1)}} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $r = \dfrac{z}{z + 5} ; \space z \neq -9 ; \space z \neq -1 $